我可以使用以下jQuery代码使用ajax请求的POST方法执行文件上载吗？
$.ajax({
type: "POST",
timeout: 50000,
url: url,
data: dataString,
success: function (data) {
alert('success');
return false;
}
});
如果可能，我需要填写数据部分吗？这是正确的方式吗？我只将文件POST到服务器端。
我一直在搜索，但我发现的是一个插件，而在我的计划中，我不想使用它。至少目前是这样。
当前回答
2019年更新：
html
<form class="fr" method='POST' enctype="multipart/form-data"> {% csrf_token %}
<textarea name='text'>
<input name='example_image'>
<button type="submit">
</form>
js
$(document).on('submit', '.fr', function(){
$.ajax({
type: 'post',
url: url, <--- you insert proper URL path to call your views.py function here.
enctype: 'multipart/form-data',
processData: false,
contentType: false,
data: new FormData(this) ,
success: function(data) {
console.log(data);
}
});
return false;
});
视图.py
form = ThisForm(request.POST, request.FILES)
if form.is_valid():
text = form.cleaned_data.get("text")
example_image = request.FILES['example_image']
其他回答
通过ajax上传文件不再需要Iframes。我最近自己做过。查看以下页面：
使用AJAX和jQuery上传HTML5文件
http://dev.w3.org/2006/webapi/FileAPI/#FileReader-界面
更新了答案并进行了清理。使用getSize函数检查大小或使用getType函数检查类型。添加了progressbar html和css代码。
var Upload = function (file) {
this.file = file;
};
Upload.prototype.getType = function() {
return this.file.type;
};
Upload.prototype.getSize = function() {
return this.file.size;
};
Upload.prototype.getName = function() {
return this.file.name;
};
Upload.prototype.doUpload = function () {
var that = this;
var formData = new FormData();
// add assoc key values, this will be posts values
formData.append("file", this.file, this.getName());
formData.append("upload_file", true);
$.ajax({
type: "POST",
url: "script",
xhr: function () {
var myXhr = $.ajaxSettings.xhr();
if (myXhr.upload) {
myXhr.upload.addEventListener('progress', that.progressHandling, false);
}
return myXhr;
},
success: function (data) {
// your callback here
},
error: function (error) {
// handle error
},
async: true,
data: formData,
cache: false,
contentType: false,
processData: false,
timeout: 60000
});
};
Upload.prototype.progressHandling = function (event) {
var percent = 0;
var position = event.loaded || event.position;
var total = event.total;
var progress_bar_id = "#progress-wrp";
if (event.lengthComputable) {
percent = Math.ceil(position / total * 100);
}
// update progressbars classes so it fits your code
$(progress_bar_id + " .progress-bar").css("width", +percent + "%");
$(progress_bar_id + " .status").text(percent + "%");
};
如何使用Upload类
//Change id to your id
$("#ingredient_file").on("change", function (e) {
var file = $(this)[0].files[0];
var upload = new Upload(file);
// maby check size or type here with upload.getSize() and upload.getType()
// execute upload
upload.doUpload();
});
Progressbar html代码
<div id="progress-wrp">
<div class="progress-bar"></div>
<div class="status">0%</div>
</div>
Progressbar css代码
#progress-wrp {
border: 1px solid #0099CC;
padding: 1px;
position: relative;
height: 30px;
border-radius: 3px;
margin: 10px;
text-align: left;
background: #fff;
box-shadow: inset 1px 3px 6px rgba(0, 0, 0, 0.12);
}
#progress-wrp .progress-bar {
height: 100%;
border-radius: 3px;
background-color: #f39ac7;
width: 0;
box-shadow: inset 1px 1px 10px rgba(0, 0, 0, 0.11);
}
#progress-wrp .status {
top: 3px;
left: 50%;
position: absolute;
display: inline-block;
color: #000000;
}
使用XMLHttpRequest（）确实可以进行AJAX上载。无需iframes。可以显示上载进度。
有关详细信息，请参阅：答案https://stackoverflow.com/a/4943774/873282对jQuery Upload Progress和AJAX文件上传提出质疑。
2019年更新：
html
<form class="fr" method='POST' enctype="multipart/form-data"> {% csrf_token %}
<textarea name='text'>
<input name='example_image'>
<button type="submit">
</form>
js
$(document).on('submit', '.fr', function(){
$.ajax({
type: 'post',
url: url, <--- you insert proper URL path to call your views.py function here.
enctype: 'multipart/form-data',
processData: false,
contentType: false,
data: new FormData(this) ,
success: function(data) {
console.log(data);
}
});
return false;
});
视图.py
form = ThisForm(request.POST, request.FILES)
if form.is_valid():
text = form.cleaned_data.get("text")
example_image = request.FILES['example_image']
var dataform = new FormData($("#myform")[0]);
//console.log(dataform);
$.ajax({
url: 'url',
type: 'POST',
data: dataform,
async: false,
success: function(res) {
response data;
},
cache: false,
contentType: false,
processData: false
});
我已经很晚了，但我正在寻找一个基于ajax的图像上传解决方案，我正在寻找的答案在这篇文章中有点分散。我确定的解决方案涉及FormData对象。我组装了一个基本形式的代码。您可以看到它演示了如何使用fd.append（）向表单添加自定义字段，以及如何在完成ajax请求时处理响应数据。
上传html:
<!DOCTYPE html>
<html>
<head>
<title>Image Upload Form</title>
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript">
function submitForm() {
console.log("submit event");
var fd = new FormData(document.getElementById("fileinfo"));
fd.append("label", "WEBUPLOAD");
$.ajax({
url: "upload.php",
type: "POST",
data: fd,
processData: false, // tell jQuery not to process the data
contentType: false // tell jQuery not to set contentType
}).done(function( data ) {
console.log("PHP Output:");
console.log( data );
});
return false;
}
</script>
</head>
<body>
<form method="post" id="fileinfo" name="fileinfo" onsubmit="return submitForm();">
<label>Select a file:</label><br>
<input type="file" name="file" required />
<input type="submit" value="Upload" />
</form>
<div id="output"></div>
</body>
</html>
如果您使用php，这里有一种处理上传的方法，包括使用上面html中演示的两个自定义字段。
上传.php
<?php
if ($_POST["label"]) {
$label = $_POST["label"];
}
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 200000)
&& in_array($extension, $allowedExts)) {
if ($_FILES["file"]["error"] > 0) {
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$filename = $label.$_FILES["file"]["name"];
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("uploads/" . $filename)) {
echo $filename . " already exists. ";
} else {
move_uploaded_file($_FILES["file"]["tmp_name"],
"uploads/" . $filename);
echo "Stored in: " . "uploads/" . $filename;
}
}
} else {
echo "Invalid file";
}
?>
